Suppose that $V$ is a vector space, $x'$ is a vector in $V$, and $y$ is a linear functional on $V$; write $Ax = [x, y]x'$ for every $x \in V$. Under what conditions on $x'$ and $y$ is $A$ a projection?
Source: Halmos, Finite dimensional vector spaces, section 43 projection exercise #2.
Below is what I think.
Since $A$ is a projection if and only if $A=A^2$. Now $Ax = [x, y]x'=y(x)x'$, $A^2(x)=A(Ax)=A(y(x)x')=[y(x)x', y]x'$. Hence, if $x=y(x)x'$, then $A$ is a projection.
Am I right? My solution seems too good to be true.
Please help, if that is not right, what's the right way to approach this problem?
No, and your condition $x=y(x)x'$ does not depend only on $A$, so it makes no sense. You should have $$A^2x=A(Ax)=A(y(x)x')=y(y(x)x')x'=y(x)y(x')x'=y(x')y(x)x'=y(x')Ax$$ so $A^2=A$ if an only if $y(x')=1$.
It is a bit more readable writing $Ax=x'.y(x)$ (so writing scalars to the right of vectors separated by a dot) so that by linearity of $A$ $$A^2x=A(x'.y(x))=A(x').y(x)=x'.y(x').y(x)=x'.y(x).y(x')=Ax.y(x').$$ Either way commutation of two scalar multiplications occurs (just for those wondering whether this can be done in general over skew fields; it cannot;-).